$\overline{AC}$ is $10$ units long $\overline{BC}$ is $8$ units long $\overline{AB}$ is $2\sqrt{41}$ units long What is $\tan(\angle ABC)$ ? $A$ $C$ $B$ $10$ $8$ $2\sqrt{41}$
Solution: SOH CAH TOA an = pposite over djacent opposite $= \overline{AC} = 10$ adjacent $= \overline{BC} = 8$ $\tan(\angle ABC)=\dfrac{10}{8}$ $=\dfrac{5}{4}$